Consider the attached file a pedigree of a rare autosomal recessive disease

(a) Choosing M  for  unaffected  and m  for  the  disorder,  male  B must   be M/m,   and  female
A  has  a  2/ 3  chance  of   being M/m.   T he  overall  chance  of   an  affected  child  is  1  *  2/ 3  *
1/ 4  =  1/ 6.

(b)  I f  C' s mother  A  is  heterozygous,  C  stands  a  1/ 2  chance  of   being  heterozygous.  D' s
mot her must   be  heterozygous,   and D  stands  a  1/ 2  chance  of   inheriting  that
heterozygosity.   The  overall  chance  of   an  affect ed  child  is  2/ 3  *  1/ 2  *  1  *  1/ 2  *  1/ 4  =
1/ 24.

(c)  The  probability  is  still  1/ 24.

(d) Now  that  we  know  individuals C  and D must   both  be M/m,   the  chance  of   the  second
child  being m/m  is  1/ 4.

(a)
Since A is unaffected, her sister is affected, and both her parents are affected, we can determine the genotypes:
Parents: both heterozygous Rr
Sister: rr
If we make a Punnett square for A, we get 1/3 RR, 2/3 Rr, and 0 rr (She's unaffected).  Therefore, A has a 2/3 chance of being a carrier.  In a mating, she has a 1/2 chance of passing either allele.  By the product rule, 2/3*1/2 = 1/3

Since B is unaffected, he can only be Rr (since his mother is affected rr and his father is unaffected).  That is, he has a 1/2 chance of passing on an affected r allele.

By the product rule, 1/3 * 1/2 = 1/6
 
(b)
Now that we have determined that there is a 2/3 probability for A to be Rr, and we know that her husband will definitely be RR (the question states that no one marrying into the family will carry the affected allele), C has a 2/3 * 1/2 = 1/3 chance to be Rr, otherwise he is RR.  D's mother must be Rr because she is unaffected and received an r allele from her own mother.  Because D's father is homozygous RR, she has a 1/2 chance of being Rr.  When two Rr people mate, there is a 1/4 chance they will have an rr offspring.  Multiplying the probabilities, 1/3 * 1/2 * 1/4, we get 1/24

(c)
Births are independent events (like the roll of a die), the affectedness/unaffectedness of one offspring will not affect the siblings.  Because a normal birth will not distinguish between RR and Rr for either parent, the chance is still 1/24.

(d)
Here, we have new information.  Because C and D are either RR or Rr and they had an affected child, we know that they MUST be Rr (otherwise they would not have been able to produce an rr child).  If we make a Punnett square for Rr x Rr, we get 1/4 RR, 1/2 Rr, and 1/4 rr.  The 1/4 rr would be the probability of the second child (and all other children they have) being affected.

(d)